After reading several questions and examples I came with this example which I modified a bit to make it work as expected.

Collections.sort(listToOrder, Comparator.comparing(item,java,list,collections"/>
	














    

		

        

            
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首页  > 解决方案 > 订单 with the order of List java

After reading several questions and examples I came with this example which I modified a bit to make it work as expected.


Collections.sort(listToOrder, Comparator.comparing(item












java - 订单 with the order of List java

After reading several questions and examples I came with this example which I modified a bit to make it work as expected.


Collections.sort(listToOrder, Comparator.comparing(item





问题描述



After reading several questions and examples I came with this example which I modified a bit to make it work as expected.


Collections.sort(listToOrder, Comparator.comparing(item -> someObject.getListOfLongs().indexOf(item.getId())));



So listToOrder is a list of MyObject which has name and id, so I need to order listToOrder in the same order as listOfLongs.


With the example of code given it work as expected however if the listToOrder is different in size it fails, wondering how I could make it to work even if the sizes are different.


Edit:
I misread, the error I was getting was an IndexOutOfBoundsException which wasn't triggered by the line of code I put up there, it was because of a manual log.









This has nothing to do with sorting, but ordering. Having the following object with full-args constructor and getters:


public static class MyObject {
    private final long id;
    private final String name;
}



... and the following data in a random order ...


List<Integer> ids = Arrays.asList(5,4,7,0,2,1,3,8,6);
List<MyObject> list = Arrays.asList(
    new MyObject(1, "one"),
    new MyObject(3, "three"),
    ...
    new MyObject(6, "six"),
    new MyObject(8, "eight")
);



The solution you are looking for is this:


List<MyObject> newList = new ArrayList<>(list);
for (int i=0; i<ids.size(); i++) {
   int id = ids.get(i);
   for (MyObject myObject: list) {
        if (myObject.getId() == id) {
            newList.set(i, myObject);
            break;
        }
    }
}



Simply find the object with the matching ID and set it to a new list. There is no dedicated method to do that.


标签： javalistcollections


解决方案









List.indexOf() returns -1 if the element is not found, which means such items will be ordered first in the resulting sorted list.


Without ordering data, the only other sensible way to handle such elements is to order them last:


Collections.sort(listToOrder, Comparator.comparing(item -> someObject.getListOfLongs().contains(item.getId()) ? someObject.getListOfLongs().indexOf(item.getId()) : Integer.MAX_VALUE));











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